Haskell Assignment
A.First Edition
This is my first edition of assignment.
Very simple.
C.Further improvement
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--Name: Qingzhe Huang
--ID: 5037735
--Title: Comp348 Assignment 1
module A1 where
--------------------------------------------------------------------------------------------------
--1
--I don't know what should I print out for an empty list except a "-1000"
--to indicate an error.
--first :: (Ord a, Num a) => a -> [a]] -> a
first e (x:[]) = if abs x < e then x else (-1000)
first e (x:y:s) = if abs (x-y) < e then x else first e (y:s)
--------------------------------------------------------------------------------------------------
--2
--(f ((a+b)/2))*(f a)>0 means that the mmiddle points has same sign as f(a),
--to do this check, we don't have to make aassumption that f(a)<0<f(b),
--because f(a)>0>f(b) can also work
--this utility function help reading
-- half :: Fractional a => a -> a -&> a
half a b = (a+b)/2
--I cannot use "real 0" to stop tthe recursive as "cos" function won't stop the infinite loop
--even f(x) is as small as 1.0e-23. So, I aarbitrarily chose 1.0e-6 to indicate 0.
-- bisect :: (Ord a, Fractional b, Fractiional a) => (b -> a) -> b -> b -> [b]
bisect f a b = (half a b):( if abs (f (half a b))< (1.0e-6) then []
else if (((f (half a b))*(f a)) >0) then bisect f (half a b) b
else if (((f (half a b))*(f a)) <0) then bisect f a (half a b) else [])
--------------------------------------------------------------------------------------------------
--3
--I really don't agree with the use of firsst here. Because mathmatically speaking it is meaningless as
--first only finds out the first consequtivve variables which difference is smaller than tolerance.
--However, it should be the function value that matters, not the variables.
{---We cannot use first here, and I have to use a self-defined function firstf
--firstf f e (x:[]) = if abs (f x) < e tthen x else 1000
--firstf f e (x:y:s) = if abs (f x- f y)<t; e then x else firstf f e (y:s)
--solve1 e f a b = firstf f e (bisect f a bb)-}
--solve1 :: (Ord a, Fractional b, Fractionaal a, Ord b) => a -> (a -> b) -> a -> a -> a
solve1 e f a b = first e (bisect f a b)
--------------------------------------------------------------------------------------------------
--4
--this is straightforward!
--diff :: Fractional a => (a -> a) -&> a -> a
diff f x = (f (x+h) - f (x-h)) /2/h where h = 0.00001
--------------------------------------------------------------------------------------------------
--5
--I have to add extra condition to prevent "divide 0" error!
--And what's more I need a small number to indicate it is zero now.
--newt :: Fractional a => (a -> a) -&> a -> [a]
newt f x = x:(if abs (f x) <1.0e-6||(diff f x==0) then [] else newt f (x-(f x)/(diff f x)))
--------------------------------------------------------------------------------------------------
--6
--solve2 :: (Ord a, Fractional a) => a --> (a -> a) -> a -> b -> a
solve2 e f a b = first e (newt f a)
--------------------------------------------------------------------------------------------------
--7
--I use 0.0001 as tolerance.
--solve :: (Ord a, Fractional a) => (a --> a) -> a -> a -> (a,a)
solve f a b= (solve1 e f a b, solve2 e f a b) where e =0.0001
--------------------------------------------------------------------------------------------------
--8
--trape :: Fractional a => (a -> a) --> a -> a -> a
trape f a b = (f a + f b)*(b-a)*0.5
--slope :: Fractional a => (a -> a) --> a -> a -> a
slope f a b = (f b - f a)/(b-a)
--integrate :: (Fractional a, Ord a) => (a -> a) -> a -> a -> a
integrate f a b = if abs (slope f a (half a b) - slope f (half a b) b) <e
then (trape f a (half a b))*2
else integrate f a (half a b) + integrate f (half a b) b where e =0.001
--integrate f a b = if abs ((trape f a (hallf a b)) - (trape f (half a b) b)) < e
-- then (trape f a (half a b))*2
-- else integrate f a (half a b) + integraate f (half a b) b where e =0.001
--------------------------------------------------------------------------------------------------
--9
--union :: Eq a => [a] -> [a] -> [[a]
union [] ys = ys
union (x:xs) ys = x:union xs (filter (x/=) ys)
--powerset :: Eq a => [a] -> [[a]]
powerset [x] = [[], [x]]
powerset (x:xs) = union (powerset xs) (map (x:) (powerset xs))
--------------------------------------------------------------------------------------------------
--10
--select :: (Eq a, Num b) => b -> [a]] -> [[a]]
select n [] = []
select 1 s = map (\x->x:[]) s
select n (x:xs) = union (map (x:) (select (n-1) xs)) (select n xs)
------------------------------------------------------------------------------------------------
--The test part:
--------------------------------------------------------------------------------------------------
test1 = first 0.01 [0.9,0.99,0.999,0.9999]
{-
A1> test1
0.99
-}
--------------------------------------------------------------------------------------------------
test2 = bisect sin 4 3
{-
A1> test2
[3.5,3.25,3.125,3.1875,3.15625,3.14063,3.14844,3.14453,3.14258,3.1416,3.14111,3.
14136,3.14148,3.14154,3.14157,3.14159,3.14159,3.14159,3.14159]
-}
--------------------------------------------------------------------------------------------------
test3 = solve1 0.0001 sin 4 3
{-
A1> test3
3.14148
-}
--------------------------------------------------------------------------------------------------
test4 = diff sin 0
{-
A1> test4
1.0
-}
--------------------------------------------------------------------------------------------------
test5 = newt sin 2
{-
A1> test5
[2.0,4.17935,2.48933,3.25214,3.14131,3.14159]
-}
--------------------------------------------------------------------------------------------------
test6 = solve2 0.001 sin 4 3
{-
A1> test6
3.1416
-}
--------------------------------------------------------------------------------------------------
test7 = solve sin 4 3
{-
A1> test7
(3.14148,3.1416)
-}
--------------------------------------------------------------------------------------------------
test8 = integrate sin 0 pi
{-
A1> test8
2.0
-}
--------------------------------------------------------------------------------------------------
test9 = powerset [1..4]
{-
A1> test9
[[],[4],[3],[3,4],[2],[2,4],[2,3],[2,3,4],[1],[1,4],[1,3],[1,3,4],[1,2],[1,2,4],
[1,2,3],[1,2,3,4]]
-}
--------------------------------------------------------------------------------------------------
test10 = select 3 [1..5]
{-
A1> test10
[[1,2,3],[1,2,4],[1,2,5],[1,3,4],[1,3,5],[1,4,5],[2,3,4],[2,3,5],[2,4,5],[3,4,5]
]
-}
--------------------------------------------------------------------------------------------------